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Math Help - Basic topology question

  1. #1
    MHF Contributor Drexel28's Avatar
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    Basic topology question

    This is relatively easy, but it requires an astute observation. You use this quite a bit later in topology so I guess it's a good challenge question.

    NOTE: If you have already seen the solution to this, please don't ruin the fun! This is aimed at people who haven't seen a lot of topology before.

    Problem: Define \mathbb{R}^{\infty} to be the set of all square-summable real sequences \{x_n\} equipped with the metric d\left(\{x_n\},\{y_n\}\right)=\sqrt{\sum_{n=1}^{\i  nfty}\left|x_n-y_n\right|^2}. Prove that this metric space is separable.

    Note: Separable means it contains a countable dense subset. Also, while my book calls this \mathbb{R}^{\infty} it is most commonly known as \ell^2.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Drexel28 View Post
    This is relatively easy, but it requires an astute observation. You use this quite a bit later in topology so I guess it's a good challenge question.

    NOTE: If you have already seen the solution to this, please don't ruin the fun! This is aimed at people who haven't seen a lot of topology before.

    Problem: Define \mathbb{R}^{\infty} to be the set of all square-summable real sequences \{x_n\} equipped with the metric d\left(\{x_n\},\{y_n\}\right)=\sqrt{\sum_{n=1}^{\i  nfty}\left|x_n-y_n\right|^2}. Prove that this metric space is separable.

    Note: Separable means it contains a countable dense subset. Also, while my book calls this \mathbb{R}^{\infty} it is most commonly known as \ell^2.

    Here is the answer. Try to prove it.

    Spoiler:


    Consider the set F of all eventually zero rational sequences

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  3. #3
    Member Black's Avatar
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    Spoiler:

    Proof: Let F be all eventually zero rational sequences

    F=\{(r_1,r_2,\dots,r_n,0,0,\dots)|r_i \in \mathbb{Q},n\in \mathbb{N}\}.

    Clearly, F \subseteq \ell^2 and is countable. I claim \text{cl}(F)=\ell^2.

    Let y=(y_k)_{k \in \mathbb{N}} \in \ell^2 and \varepsilon>0. Pick n_0 \in \mathbb{N} such that \sum_{k=n_0+1}^{\infty}|y_k|^2 =M_{n_0} < \varepsilon^2. Since \mathbb{Q} is dense in \mathbb{R}, for each y_k (for 1\le k \le n_0), pick r_{0k} \in \mathbb{Q} such that

    |r_{0k}-y_k| < \sqrt{\frac{1}{n_0}\left(\varepsilon^2-M_{n_0}\right)} \Longrightarrow |r_{0k}-y_k|^2 < \frac{1}{n_0}\left(\varepsilon^2-M_{n_0}\right).

    So we have

    \sum_{k=1}^{n_0}|r_{0k}-y_k|^2<\varepsilon^2-M_{n_0}.

    If we let x_0 = (r_{01},r_{02},\dots,r_{0n_0},0,0,\dots) \in F, then d(x_0,y)<\varepsilon. Therefore, y is a limit point of F.
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